Another Tough Nut! Take any rectangle ABCD such that AB > BC and say the lengths of AB and CD are S and s respectively. The point P is on AB and Q is on CD. For APCQ to be a rhombus, the lengths AP and PC must be equal. Consider the point P coinciding with A (such that AP=0) and then P moving along AB so that the length AP increases continuously from 0 to S while the length of PC decreases continuously from

______ ÖS2 + s2

to s . As AP < PC initially (when P is at A) and AP > PC finally (when P is at B) there must be one point at which AP = PC. Similarly there is exactly one position of Q such that CQ = QA making APCQ into a rhombus.

Now take AP = PC = x than you can use Pythagoras Theorem to find x in terms of S and s so that you can find the ratio of the areas of the areas of the rhombus and the rectangle.